QuestionEvaluate the integral $\int_{0}^{\pi / 6} \frac{\sin t}{\cos ^{2} t} d t$ .

Studdy Solution

STEP 1

Assumptions1. We are asked to evaluate the definite integral $\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{} t} d t$ . We know the integral and trigonometric identities

STEP 2

The integral can be simplified by using a substitution. We can let $u = \cos t$ , so $du = -\sin t\, dt$ .
$\int \frac{\sin t}{\cos ^{2} t} d t = -\int \frac{1}{u^{2}} du$

STEP 3

Now, we can evaluate the integral using the power rule for integration, which states that $\int x^n dx = \frac{1}{n+1}x^{n+1}$ .
$-\int \frac{1}{u^{2}} du = -\int u^{-2} du = \frac{1}{-1+1}u^{-1+1} = u^{-1}$

STEP 4

Substitute $u = \cos t$ back into the integral.
$u^{-1} = \frac{1}{\cos t}$

STEP 5

Now we need to evaluate the definite integral from $0$ to $\pi /$ . To do this, we substitute these values into the integral.
$\left[\frac{1}{\cos t}\right]_{0}^{\pi /} = \frac{1}{\cos(\pi /)} - \frac{1}{\cos(0)}$

STEP 6

Evaluate the cosine values.
$\frac{1}{\cos(\pi /6)} - \frac{1}{\cos(0)} = \frac{1}{\sqrt{3}/2} - \frac{1}{1}$

STEP 7

implify the fractions.
$\frac{1}{\sqrt{3}/2} - \frac{1}{1} =2\sqrt{3} -1$ So, the solution to the integral $\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{2} t} d t$ is $2\sqrt{3} -1$ .

Was this helpful?

QuestionEvaluate the integral $\int_{0}^{\pi / 6} \frac{\sin t}{\cos ^{2} t} d t$ .

Studdy Solution

STEP 1

Assumptions1. We are asked to evaluate the definite integral $\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{} t} d t$ . We know the integral and trigonometric identities

STEP 2

The integral can be simplified by using a substitution. We can let $u = \cos t$ , so $du = -\sin t\, dt$ .
$\int \frac{\sin t}{\cos ^{2} t} d t = -\int \frac{1}{u^{2}} du$

STEP 3

Now, we can evaluate the integral using the power rule for integration, which states that $\int x^n dx = \frac{1}{n+1}x^{n+1}$ .
$-\int \frac{1}{u^{2}} du = -\int u^{-2} du = \frac{1}{-1+1}u^{-1+1} = u^{-1}$

STEP 4

Substitute $u = \cos t$ back into the integral.
$u^{-1} = \frac{1}{\cos t}$

STEP 5

Now we need to evaluate the definite integral from $0$ to $\pi /$ . To do this, we substitute these values into the integral.
$\left[\frac{1}{\cos t}\right]_{0}^{\pi /} = \frac{1}{\cos(\pi /)} - \frac{1}{\cos(0)}$

STEP 6

Evaluate the cosine values.
$\frac{1}{\cos(\pi /6)} - \frac{1}{\cos(0)} = \frac{1}{\sqrt{3}/2} - \frac{1}{1}$

STEP 7

implify the fractions.
$\frac{1}{\sqrt{3}/2} - \frac{1}{1} =2\sqrt{3} -1$ So, the solution to the integral $\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{2} t} d t$ is $2\sqrt{3} -1$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionEvaluate the integral ∫0π/6sin⁡tcos⁡2tdt\int_{0}^{\pi / 6} \frac{\sin t}{\cos ^{2} t} d t∫0π/6​cos2tsint​dt.

Studdy Solution

STEP 1

Assumptions1. We are asked to evaluate the definite integral ∫0π/6sin⁡tcos⁡tdt\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{} t} d t∫0π/6​costsint​dt . We know the integral and trigonometric identities

STEP 2

The integral can be simplified by using a substitution. We can let u=cos⁡tu = \cos tu=cost, so du=−sin⁡t dtdu = -\sin t\, dtdu=−sintdt.∫sin⁡tcos⁡2tdt=−∫1u2du\int \frac{\sin t}{\cos ^{2} t} d t = -\int \frac{1}{u^{2}} du∫cos2tsint​dt=−∫u21​du

STEP 3

STEP 4

Substitute u=cos⁡tu = \cos tu=cost back into the integral.u−1=1cos⁡tu^{-1} = \frac{1}{\cos t}u−1=cost1​

STEP 5

Now we need to evaluate the definite integral from 000 to π/\pi /π/. To do this, we substitute these values into the integral.[1cos⁡t]0π/=1cos⁡(π/)−1cos⁡(0)\left[\frac{1}{\cos t}\right]_{0}^{\pi /} = \frac{1}{\cos(\pi /)} - \frac{1}{\cos(0)}[cost1​]0π/​=cos(π/)1​−cos(0)1​

STEP 6

Evaluate the cosine values.1cos⁡(π/6)−1cos⁡(0)=13/2−11\frac{1}{\cos(\pi /6)} - \frac{1}{\cos(0)} = \frac{1}{\sqrt{3}/2} - \frac{1}{1}cos(π/6)1​−cos(0)1​=3​/21​−11​

STEP 7

implify the fractions.13/2−11=23−1\frac{1}{\sqrt{3}/2} - \frac{1}{1} =2\sqrt{3} -13​/21​−11​=23​−1So, the solution to the integral ∫0π/6sin⁡tcos⁡2tdt\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{2} t} d t∫0π/6​cos2tsint​dt is 23−12\sqrt{3} -123​−1.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionEvaluate the integral $\int_{0}^{\pi / 6} \frac{\sin t}{\cos ^{2} t} d t$ .

Assumptions1. We are asked to evaluate the definite integral $\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{} t} d t$ . We know the integral and trigonometric identities

The integral can be simplified by using a substitution. We can let $u = \cos t$ , so $du = -\sin t\, dt$ .
$\int \frac{\sin t}{\cos ^{2} t} d t = -\int \frac{1}{u^{2}} du$

Substitute $u = \cos t$ back into the integral.
$u^{-1} = \frac{1}{\cos t}$

Now we need to evaluate the definite integral from $0$ to $\pi /$ . To do this, we substitute these values into the integral.
$\left[\frac{1}{\cos t}\right]_{0}^{\pi /} = \frac{1}{\cos(\pi /)} - \frac{1}{\cos(0)}$

Evaluate the cosine values.
$\frac{1}{\cos(\pi /6)} - \frac{1}{\cos(0)} = \frac{1}{\sqrt{3}/2} - \frac{1}{1}$

implify the fractions.
$\frac{1}{\sqrt{3}/2} - \frac{1}{1} =2\sqrt{3} -1$ So, the solution to the integral $\int_{0}^{\pi /6} \frac{\sin t}{\cos ^{2} t} d t$ is $2\sqrt{3} -1$ .